题目
Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.
A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:
- 0 <= i, j < nums.length
- i != j
- a <= b
- b - a == k
解题思路
使用计数器统计每个数出现次数，如果k=0，那么结果是有多少个数出现过2次以上，否则，遍历每个数n，判断n+k是否出现过
代码
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
ct = Counter(nums)
res = 0
if k == 0:
for v in ct.values():
res += v > 1
else:
for n in ct:
res += k + n in ct
return res
